Let 'l' is the length of median from the vertex A to the side BC of a `Delta ABC`. Then

Options

A. `4l^(2)=2b^(2)+2c^(2)-a^(2)`
B. `4l^(2)=b^(2)+c^(2)+2bc cos A`
C. `4l^(2)=a^(2)+4bc cos A`
D. `4l^(2)=(2s-a)^(2)"sin"^(2)(A)/(2)`

Explanation

To find the length of the median \( l \) from vertex \( A \) to side \( BC \) in triangle \( ABC \), we can use the Apollonius theorem. The theorem states that:

\[ AB^2 + AC^2 = 2AD^2 + 2BD^2 \]

where \( D \) is the midpoint of side \( BC \).

### Step-by-step Solution:

1. **Identify the sides of the triangle**:
Let \( AB = c \), \( AC = b \), and \( BC = a \). Since \( D \) is the midpoint of \( BC \), we have \( BD = DC = \frac{a}{2} \).

2. **Apply the Apollonius theorem**:
According to the theorem, we can write:
\[ c^2 + b^2 = 2l^2 + 2\left(\frac{a}{2}\right)^2 \]
Simplifying the right side:
\[ c^2 + b^2 = 2l^2 + 2\left(\frac{a^2}{4}\right) = 2l^2 + \frac{a^2}{2} \]

3. **Rearranging the equation**:
Rearranging gives us:
\[ 2l^2 = c^2 + b^2 - \frac{a^2}{2} \]

4. **Divide by 2**:
Dividing both sides by 2 to isolate \( l^2 \):
\[ l^2 = \frac{c^2 + b^2 - \frac{a^2}{2}}{2} \]

5. **Final expression for \( l \)**:
Taking the square root gives us:
\[ l = \sqrt{\frac{c^2 + b^2 - \frac{a^2}{2}}{2}} \]

### Conclusion:
Thus, the length of the median \( l \) from vertex \( A \) to side \( BC \) is given by:
\[ l = \sqrt{\frac{b^2 + c^2 - \frac{a^2}{2}}{2}} \]